This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 1: Apply the principle of conservation of mechanical energy. Since the system is released from rest and the pulley is frictionless, mechanical energy is conserved. The initial potential energy of M_2 is converted into kinetic energy of both masses and potential energy gained by M_1. The initial total energy (E_i) is the potential energy of M_2 (assuming M_1 starts at a reference height of 0). E_i = M_2 g h_2 The final total energy (E_f) is the potential energy of M_1 (which has moved up by h_2) plus the kinetic energy of both masses just before M_2 hits the ground. E_f = M_1 g h_2 + (1)/(2) M_1 v_f^2 + (1)/(2) M_2 v_f^2 By conservation of energy, E_i = E_f. M_2 g h_2 = M_1 g h_2 + (1)/(2) M_1 v_f^2 + (1)/(2) M_2 v_f^2 Step 2: Rearrange the equation to solve for the final velocity v_f. M_2 g h_2 - M_1 g h_2 = (1)/(2) (M_1 + M_2) v_f^2 g h_2 (M_2 - M_1) = (1)/(2) (M_1 + M_2) v_f^2 v_f^2 = (2 g h_2 (M_2 - M_1))/(M_1 + M_2) v_f = sqrt((2 g h_2 (M_2 - M_1))/(M_1 + M_2)) Step 3: Substitute the given values into the equation. Given: M_1 = 1.5 kg M_2 = 4.5 kg h_2 = 2 m g = 9.81 m/s^2 v_f = sqrt(2 × 9.81 m/s)^2 × 2 m × (4.5 kg - 1.5 kg)1.5 kg + 4.5 kg v_f = sqrt((2 × 9.81 × 2 × (3.0))/(6.0)) m/s v_f = sqrt((117.72)/(6.0)) m/s v_f = sqrt(19.62) m/s v_f ≈ 4.429 m/s The velocity with which M_2 hits the ground is approximately 4.43 m/s. 4.43 m/s That's 2 down. 3 left today — send the next one.